Question

Three persons $A,B$ and $C$ are to speak at a function along with $7$ other person. The probability that $A$ speaks before $B$ and $B$ speaks before $C$ is

• A. $3/7$
• B. $1/6$
• C. $4/7$
• D. None of these

Answer 1

Answer: (B)

$1/6$

The total number of ways in which $10$ persons can speak is ${}^{10}{P}_{10}=10!$

The number of ways in which $A,B$ and $C$ can be arranged in the specefied speaking order is ${}^{10}{C}_3$.

There are $7!$ ways in which other seven can speak.

So, the favorable number of ways is ${}^{10}{C}_3$

Hence, the probability $=\frac{{}^{10}{C}_3\times 7!}{10!}=\frac{1}{6}$